Problem: Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())
#define ll long long
#define pb(a) push_back(a)
#define INF 1000000000
#define LINF 3e18+7
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

typedef vector<unsigned int>vec;
typedef vector<vec> mat;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const double PI = acos(-1.0);
const double EPS = 1e-10;

struct RollingHash {
    typedef long long int_type;
    typedef pair<int_type,int_type> hash_type;
 
    int_type base1;
    int_type base2;
    int_type mod1;
    int_type mod2;
    vector<int_type> hash1;
    vector<int_type> hash2;
    vector<int_type> pow1;
    vector<int_type> pow2;
 
    void init(const string &s) {
        int n = s.size();
 
        hash1.assign(n+1,0);
        hash2.assign(n+1,0);
        pow1.assign(n+1,1);
        pow2.assign(n+1,1);
 
        for(int i=0;i<n;i++) {
            hash1[i+1] = (hash1[i]+s[i]) * base1 % mod1;
            hash2[i+1] = (hash2[i]+s[i]) * base2 % mod2;
            pow1[i+1] = pow1[i] * base1 % mod1;
            pow2[i+1] = pow2[i] * base2 % mod2;
        }
    }
 
    RollingHash() : base1(1009), base2(1007), mod1(1000000007), mod2(1000000009) {}
 
    hash_type get(int l,int r) {
        int_type t1 = ((hash1[r] - hash1[l] * pow1[r-l]) % mod1 + mod1) % mod1;
        int_type t2 = ((hash2[r] - hash2[l] * pow2[r-l]) % mod2 + mod2) % mod2;
        return make_pair(t1, t2);
    }
 
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m;
    string s;
    cin >> n >> m >> s;
    set<RollingHash::hash_type> S;
    RollingHash rh;
    rh.init(s);

    int l = 0, r = 1;
    REP(i,m){
        string q;
        cin >> q;
        if(q == "R++") r++;
        if(q == "R--") r--;
        if(q == "L++") l++;
        if(q == "L--") l--;
        S.insert(rh.get(l,r));
    }
    cout << S.size() << endl;
}